Solve a nonlinear equation with constrains

Hello,

clear 

p0=1000e6;
t0 = 1e-6;
td = 1e-6;
t  = t0 + td;
c  = 5e6;

a  = @(r)log(r)./(t0*(r-1.0));
b  = @(r)a(r).*r;

func= @(r) p0*((exp(-a*t) - exp(-b*t))/(exp(-a*t0) - exp(-b*t0)))-c;
r=1.5;

roots = fzero(func,r)

my equation is the func.

where a and b are,

[Bora Eryilmaz]

Your function "p(t)" (func in your code) is not a function of time since you are assigning a fixed scalar value to "t" in your code. So, func() is a function of r, with a fixed "t". So what you are really solving here is p(r) = 0 given fixed values for t, t0, a, b, c.

You will need to reformulate your problem.

Answer 1

[Torsten]

p0 = 1000e6;
t0 = 1e-6;
td = 1e-6;
t  = t0 + td;
c  = 7.5e8;
a  = @(r)log(r)./(t0*(r-1.0));
b  = @(r)a(r).*r;
func= @(r) p0*((exp(-a(r)*t) - exp(-b(r)*t))./(exp(-a(r)*t0) - exp(-b(r)*t0))) - c;
r = 0.001:0.1:10;
plot(r,func(r))

root1 = fzero(func,[0.001 0.75])

root1 = 0.5000

root2 = fzero(func,[1.1 2.5])

root2 = 2.0000

[Miraboreasu]

Thanks, but if I keep c=5e6, it won't work

[Torsten]

Yes, because no roots exist. Plot the function, and you will see that it does not cross the r-axis.

Note for functions that only depend on one variable: First plot, then solve.